What is the directrix (x+3)^2=-8(y+2) ?

The directrix (x+3)^2=-8(y+2) is y=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

directrix (x+3)^2=-8(y+2)

directrix

(x + 3)^{2} = - 8 (y + 2)

y = 0

Solution steps

Rewrite

(x + 3)^{2} = - 8 (y + 2)

in the standard form:

4 (- 2) (y - (- 2)) = (x - (- 3))^{2}

(h, k) = (- 3, - 2), p = - 2

Parabola is symmetric around the y-axis and so the directrix is a line parallel to the x-axis, a distance

- p

from the center

(- 3, - 2)

y-coordinate

y = - 2 - p

y = - 2 - (- 2)

Refine

y = 0

y = 3 x^{2} + 4

y = 2 x^{2} - 8 x + 4

3 - t^{2}

y = - 8 (x + 1)^{2}

- 8 (y + 1) = (x - 2)^{2}